Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $p = \dfrac{x + 4}{-3x^2 + 12x + 135} \times \dfrac{x^2 - x - 30}{-5x + 30} $
First factor out any common factors. $p = \dfrac{x + 4}{-3(x^2 - 4x - 45)} \times \dfrac{x^2 - x - 30}{-5(x - 6)} $ Then factor the quadratic expressions. $p = \dfrac {x + 4} {-3(x + 5)(x - 9)} \times \dfrac {(x + 5)(x - 6)} {-5(x - 6)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {(x + 4) \times (x + 5)(x - 6) } { -3(x + 5)(x - 9) \times -5(x - 6)} $ $p = \dfrac {(x + 5)(x - 6)(x + 4)} {15(x + 5)(x - 9)(x - 6)} $ Notice that $(x + 5)$ and $(x - 6)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {\cancel{(x + 5)}(x - 6)(x + 4)} {15\cancel{(x + 5)}(x - 9)(x - 6)} $ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ $p = \dfrac {\cancel{(x + 5)}\cancel{(x - 6)}(x + 4)} {15\cancel{(x + 5)}(x - 9)\cancel{(x - 6)}} $ We are dividing by $x - 6$ , so $x - 6 \neq 0$ Therefore, $x \neq 6$ $p = \dfrac {x + 4} {15(x - 9)} $ $ p = \dfrac{x + 4}{15(x - 9)}; x \neq -5; x \neq 6 $